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前置知识
好像没啥，看得懂公式就行了。
引入
一般的生成函数会和某个数列 \(a_0,a_1,\dots,a_n\) 建立关系，它是一个多项式函数，其中每一项的系数等于对应数列每一位的值，这样的函数有时候也叫形式幂级数。"><meta property="og:type" content="article"><meta property="og:title" content="生成函数"><meta property="og:url" content="https://schtonn.github.io/blog/posts/gen-func/index.html"><meta property="og:site_name" content="schtonn"><meta property="og:description" content="。。。我最近写数学越来越多了，人都麻了
前置知识
好像没啥，看得懂公式就行了。
引入
一般的生成函数会和某个数列 \(a_0,a_1,\dots,a_n\) 建立关系，它是一个多项式函数，其中每一项的系数等于对应数列每一位的值，这样的函数有时候也叫形式幂级数。"><meta property="og:locale" content="en_US"><meta property="article:published_time" content="2021-02-16T13:38:17.238Z"><meta property="article:modified_time" content="2022-10-19T15:02:06.668Z"><meta property="article:author" content="Alex"><meta property="article:tag" content="math"><meta name="twitter:card" content="summary"><link rel="canonical" href="https://schtonn.github.io/blog/posts/gen-func/"><script id="page-configurations">CONFIG.page={sidebar:"",isHome:!1,isPost:!0,lang:"en"}</script><title>生成函数 | schtonn</title><noscript><style>.sidebar-inner,.use-motion .brand,.use-motion .collection-header,.use-motion .comments,.use-motion .menu-item,.use-motion .pagination,.use-motion .post-block,.use-motion .post-body,.use-motion .post-header{opacity:initial}.use-motion .site-subtitle,.use-motion .site-title{opacity:initial;top:initial}.use-motion .logo-line-before i{left:initial}.use-motion 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itemprop="description" content="blog"></span><script type="text/javascript" src="/blog/js/md5.js"></script><script></script><script>document.oncopy=function(e){window.event&&(e=window.event);try{var t=e.srcElement;return"INPUT"==t.tagName&&"text"==t.type.toLowerCase()||"TEXTAREA"==t.tagName}catch(e){return!1}}</script><span hidden itemprop="publisher" itemscope itemtype="http://schema.org/Organization"><meta itemprop="name" content="schtonn"></span><header class="post-header"><h1 class="post-title" itemprop="name headline"> 生成函数</h1><div class="post-meta"><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-calendar-o"></i></span> <span class="post-meta-item-text">Posted on</span> <time title="Created: 2021-Feb-16 21:38:17" itemprop="dateCreated datePublished" datetime="2021-02-16T21:38:17+08:00">2021-Feb-16</time></span><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-calendar-check-o"></i></span> <span class="post-meta-item-text">Edited on</span> <time title="Modified: 2022-Oct-19 23:02:06" itemprop="dateModified" datetime="2022-10-19T23:02:06+08:00">2022-Oct-19</time></span><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-comment-o"></i></span> <span class="post-meta-item-text">Valine:</span><a title="valine" href="/blog/posts/gen-func/#valine-comments" itemprop="discussionUrl"><span class="post-comments-count valine-comment-count" data-xid="/blog/posts/gen-func/" itemprop="commentCount"></span></a></span></div></header><div class="post-body" itemprop="articleBody"><p>。。。我最近写数学越来越多了，人都麻了</p><h2 id="前置知识">前置知识</h2><p>好像没啥，看得懂公式就行了。</p><h2 id="引入">引入</h2><p>一般的生成函数会和某个数列 <span class="math inline">\(a_0,a_1,\dots,a_n\)</span> 建立关系，它是一个多项式函数，其中每一项的系数等于对应数列每一位的值，这样的函数有时候也叫形式幂级数。</p><a id="more"></a><p>最常见的形式（普通生成函数）是这样的：</p><p><span class="math display">\[g(x)=\sum^\infin_{i=0}a_ix^i\]</span></p><h2 id="用于组合数学">用于组合数学</h2><p>当然，有时用在组合数学时会把生成函数缩减到一个有限项多项式，但无限项多项式不一定求不出有意义的解。</p><p>先来看一个简单的计数问题：</p><p>有一个苹果，一只香蕉和一个菠萝，求：</p><ol type="1"><li>不选有几种选法</li><li>选一个有几种选法</li><li>选两个有几种选法</li><li>选三个有几种选法</li></ol><p>平时解这类问题时，我们一般都用组合数，上述问题直接对应了 <span class="math inline">\(\dbinom{3}{0}\)</span>，<span class="math inline">\(\dbinom{3}{1}\)</span>，<span class="math inline">\(\dbinom{3}{2}\)</span> 和 <span class="math inline">\(\dbinom{3}{3}\)</span>。但这里要用一种不太一样的方式，以此更好的理解生成函数。</p><p>我们用 <span class="math inline">\(1\)</span> 代表不选，<span class="math inline">\(x\)</span> 代表选一个，<span class="math inline">\(x^2\)</span> 代表选两个，etc.</p><p>对于每种水果，都有选一个和不选两种可能，也就是 <span class="math inline">\((x+1)\)</span>。</p><p>那么对于三种水果，就可以列一个这样的式子：</p><p><span class="math display">\[\begin{array}{rl} &(x+1)(x+1)(x+1)\\ =&1+3x+3x^2+x^3 \end{array}\]</span></p><p>发现每一项的系数刚好对应了上面问题的答案。</p><p>所以 <span class="math inline">\(g(x)=(x+1)(x+1)(x+1)\)</span> 是 <span class="math inline">\(\dbinom{3}{0},\dbinom{3}{1},\dbinom{3}{2},\dbinom{3}{3}\)</span> 这个数列的生成函数。</p><p>对于更大的范围它同样成立：</p><p><span class="math display">\[g(x)=(x+1)^n\rightarrow\dbinom{n}{0},\dbinom{n}{1},\cdots,\dbinom{n}{n}\]</span></p><p>还有更神奇的应用：</p><p><span class="math display">\[\prod_{i=1}^n(1+c_ix)\]</span></p><p><span class="math display">\[=1+(c_1+\cdots+c_n)x+(c_1c_2+c_1c_3+\cdots+c_{n-1}c_n)x^2+\cdots+(c_1c_2\cdots c_n)x^n\]</span></p><p>则 <span class="math inline">\(x^i\)</span> 项的系数就是从 <span class="math inline">\(n\)</span> 个元素中选取 <span class="math inline">\(i\)</span> 个进行组合的全体。令全体 <span class="math inline">\(c_i=1\)</span>，就是上面的结论。再令 <span class="math inline">\(x=1\)</span>，得到 <span class="math inline">\(\sum\limits_{i=0}^n\dbinom{n}{i}=(1+1)^n=2^n\)</span>。</p><p>有人可能发现这完全就是二项式定理，但是它可以拓展到更大的范围。</p><h3 id="样例一">样例一</h3><p><span class="math display">\[c_1+c_2+c_3+c_4+c_5=30\]</span> 其中： <span class="math display">\[2\leq c_1\leq4,3\leq c_i\leq8(2\leq i \leq5)\]</span> 解法： <span class="math display">\[ (x^2+x^3+x^4)(x^3+x^4+x^5+x^6+x^7+x^8)^4 \]</span> 找到其中 <span class="math inline">\(x^{30}\)</span> 的系数即可。</p><h3 id="样例二">样例二</h3><p>依然是 <span class="math display">\[c_1+c_2+c_3+c_4+c_5=30\]</span> 其中： <span class="math display">\[0\leq c_i(1\leq i \leq5),c_2\in\mathbb{E},c_3\in\mathbb{O}\]</span> （<span class="math inline">\(c_2\)</span> 是偶数，<span class="math inline">\(c_3\)</span> 是奇数，注意没有上界，从这里开始接触到形式幂级数的本质了）</p><p>解法： <span class="math display">\[ (1+x^1+x^2\cdots)^3(1+x^2+x^4+\cdots)(x+x^3+x^5+\cdots) \]</span></p><p>同样地，找到 <span class="math inline">\(x^{30}\)</span> 的系数。</p><h2 id="用于数列通项公式">用于数列通项公式</h2><h3 id="特殊形式">特殊形式</h3><p>先来看一个最简单的序列：</p><p><span class="math display">\[a=1,1,1,1,1\]</span></p><p>也就是五个一。</p><p>它的生成函数就是：</p><p><span class="math display">\[g(x)=1+x+x^2+x^3+x^4\]</span></p><p>经过等比数列求和的转化可以变为：</p><p><span class="math display">\[g(x)=\dfrac{1-x^5}{1-x}\]</span></p><hr><p>如果 <span class="math inline">\(a\)</span> 有无限项呢？</p><p><span class="math display">\[a=1,1,1,\cdots\]</span></p><p><span class="math display">\[g(x)=1+x+x^2+\cdots\]</span></p><p><span class="math display">\[\sum_{i=0}^nx^i=\dfrac{1-x^n}{1-x}\]</span></p><p><span class="math display">\[n\to\infin\]</span></p><p>当 <span class="math inline">\(x\in(1,-1)\)</span> 时</p><p><span class="math display">\[x^n\to 0\]</span></p><p><span class="math display">\[\therefore g(x)=\dfrac{1}{1-x}\]</span></p><hr><p>你可能疑惑为什么能直接给 <span class="math inline">\(x\)</span> 限定域，但是上面提到了，生成函数是<em>形式幂级数</em>，<span class="math inline">\(x\)</span> 的值没有意义。从这里开始对于没有上界的生成函数，出现了一种能大大简化计算的表达方式。</p><p><span class="math display">\[g(x)=\sum_{i=0}^\infin x^i=\dfrac{1}{1-x}\]</span></p><hr><p>将 <span class="math inline">\(x\)</span> 改为 <span class="math inline">\(x^2\)</span> 得到：</p><p><span class="math display">\[g(x)=1+x^2+x^4+\cdots=\sum_{i=0}^\infin x^{2i}=\dfrac{1}{1-x^2}\]</span></p><p>将 <span class="math inline">\(x\)</span> 改为 <span class="math inline">\(x^k\)</span> 得到：</p><p><span class="math display">\[g(x)=\sum_{i=0}^\infin x^{ki}=\dfrac{1}{1-x^k}\]</span></p><p>所以 <span class="math inline">\(1,0,1,0,1,\cdots\)</span> 的生成函数是 <span class="math inline">\(\dfrac{1}{1-x^2}\)</span>。</p><hr><p>求以下数列的生成函数：</p><p><span class="math display">\[\dbinom{1}{k},\dbinom{2}{k},\dbinom{3}{k},\cdots\]</span></p><p><span class="math display">\[g(x)=\dbinom{1}{k}+\dbinom{2}{k}x+\dbinom{3}{k}x^2+\cdots\]</span></p><p><span class="math display">\[g(x)=\sum_{i=0}^\infin\dbinom{i+1}{k}x^i\]</span></p><p>运用广义二项式定理可得：</p><p><span class="math display">\[g(x)=(1+x^2+x^3+\cdots)^k=\dfrac{1}{(1-x)^k}\]</span></p><hr><p>两个生成函数相加是什么呢？</p><p>设：</p><p><span class="math display">\[g_a(x)=\sum_{i=0}^\infin a_ix^i\]</span></p><p><span class="math display">\[g_b(x)=\sum_{i=0}^\infin b_ix^i\]</span></p><p><span class="math display">\[g_a(x)+g_b(x)=\sum_{i=0}^\infin(a_i+b_i)x^i\]</span></p><p>好像没什么可解释的。这就是数列的每一项分别相加。</p><hr><p>两个生成函数相乘是什么呢？</p><p>设：</p><p><span class="math display">\[g_a(x)=\sum_{i=0}^\infin a_ix^i\]</span></p><p><span class="math display">\[g_b(x)=\sum_{i=0}^\infin b_ix^i\]</span></p><p>那么它们相乘的结果就是它们的卷积：</p><p><span class="math display">\[g_a(x)g_b(x)=\sum_{i=0}^\infin(\sum_{k=0}^ia_kb_{i-k})x^i\]</span></p><p>你可能觉得这和卷积毫无关系，但是仔细想想，这就是多项式相乘啊，做快速傅里叶变换的时候不就是加速的卷积吗？</p><hr><p>知道了这几种特殊形式，就可以快速解出很多问题，比如著名的 <a href="https://www.luogu.com.cn/problem/P2000" target="_blank" rel="noopener">Luogu P2000</a>。你也可以用这个方式快速解出上面的<a href="#样例二">样例二</a>。</p><p><span class="math display">\[g(x)=(1+x^1+x^2\cdots)^3(1+x^2+x^4+\cdots)(x+x^3+x^5+\cdots)\]</span></p><p><span class="math display">\[g(x)=\left(\dfrac{1}{1-x}\right)^3\times\dfrac{1}{1-x^2}\times\dfrac{x}{1-x^2}\]</span></p><h3 id="斐波那契数列">斐波那契数列</h3><p>很久之前我曾经发过一个<a href="/blog/posts/fibonacci">快速求斐波那契</a>的通项公式，实际上它是从生成函数来的。具体看看过程：</p><p>设 <span class="math inline">\(F_0=0\)</span>，<span class="math inline">\(F_1=1\)</span>。</p><p><span class="math display">\[g(x)=x+x^2+2x^3+3x^4+5x^5+\cdots\]</span></p><p>将生成函数乘上 <span class="math inline">\(x\)</span>，就相当于将其整体后移一位。又因为数列每一项都是前面两项的和（除了初始值不是，它会留下一个富余的 <span class="math inline">\(x\)</span>），所以：</p><p><span class="math display">\[g(x)-xg(x)-x^2g(x)=x\]</span></p><p>解得：</p><p><span class="math display">\[g(x)=\dfrac{x}{1-x-x^2}\]</span></p><p>因式分解得到：</p><p><span class="math display">\[g(x)=\dfrac{x}{\left(1-\dfrac{1-\sqrt{5}}{2}x\right)\left(1-\dfrac{1+\sqrt{5}}{2}x\right)}\]</span></p><p>运用裂项法得到：</p><p><span class="math display">\[g(x)=-\dfrac{1}{\sqrt5}\dfrac{1}{\left(1-\dfrac{1-\sqrt5}{2}x\right)} + \dfrac{1}{\sqrt5}\dfrac{1}{\left(1-\dfrac{1+\sqrt5}{2}x\right)}\]</span></p><p>可见它是两个类似上面等比数列求和 <span class="math inline">\(\dfrac{1}{1-x}\)</span> 的形式，然后乘上常数并求和。</p><p>反向运用上面的方式将其还原成数列，得到：</p><p><span class="math display">\[\operatorname{F}(n)=\dfrac{\sqrt{5}}{5}\left[\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n\right]\]</span></p><p>理论上，这种方法可以应用于任何线性齐次递推中，比如：</p><p><span class="math display">\[\begin{array}{rcl} F(n,1)&=&F(n-1,2)+2F(n-3,1)+5,\\ F(n,2)&=&F(n-1,1)+3F(n-3,1)+2F(n-3,2)+3 \end{array}\]</span></p><p>但是由于它过于复杂，我不会推。</p></div><div><ul class="post-copyright"><li class="post-copyright-author"> <strong>Post author:</strong> Alex</li><li class="post-copyright-link"> <strong>Post link:</strong> <a href="https://schtonn.github.io/blog/posts/gen-func/" title="生成函数">https://schtonn.github.io/blog/posts/gen-func/</a></li><li class="post-copyright-license"> <strong>Copyright Notice:</strong> All articles in this blog are licensed under<a href="https://creativecommons.org/licenses/by-nc-sa/4.0/" rel="noopener" target="_blank"><i class="fa fa-fw fa-creative-commons"></i> BY-NC-SA</a> unless stating additionally.</li></ul></div><footer class="post-footer"><div 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